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Ecuatie in numere naturale - JBTST III 2007, problema 2

 
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Laurian Filip
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Joined: 25 Nov 2007
Posts: 473
Location: Bucuresti

PostPosted: Tue Apr 08, 2008 6:14 pm    Post subject: Ecuatie in numere naturale - JBTST III 2007, problema 2 Reply with quote

Sa se rezolve in numere naturale ecuatia

(x^{2}+2)(y^{2}+3)(z^{2}+4)=60xyz.
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Mateescu Constantin
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Joined: 21 Apr 2009
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PostPosted: Thu Jun 25, 2009 12:15 pm    Post subject: Reply with quote

Remarcam, mai intai, ca ecuatia (x^2+2)(y^2+3)(z^2+4)=60xyz\ (1), daca are solutiii in \mathbb{N}, atunci x,\ y,\ z\in\mathbb{N}^{\ast}

Avem relatiile echivalente: \left\{\begin x^2+2\ge 3x\ \Longleftrightarrow (x-1)(x-2)\ge 0,\ \mbox{cu egalitate daca}\ \ x\in\{1,\ 2\}\ (2) \\y^2+3\ge 4y\ \Longleftrightarrow (y-1)(y-3)\ge 0,\ \mbox{cu egalitate daca}\ y\in\{1,\ 3\}\ (3) \\z^2+4\ge 5z\ \Longleftrightarrow (z-1)(z-4)\ge 0,\ \mbox{cu egalitate daca}\ z\in\{1,\ 4\}\ \ \ (4)

Din (2), (3) si (4) se observa ca (x^2+2)(y^2+3)(z^2+4)\ge 60xyz, iar daca x\in\{1,\ 3\},\ y\in\{1,\ 4\},\ z\in\{1,\ 4\} avem egalitate.

Deci ecuatia are solutiile: (x,\ y,\ z)\in\{(1,\ 1,\ 1);\ (1,\ 1,\ 4);\ (1,\ 3,\ 1);\ (1,\ 3\ ,4);\ (2,\ 1,\ 1);\ (2,\ 1,\ 4);\ (2,\ 3,\ 4);\ (2,\ 3,\ 1)\}

Aratam ca ecuatia (1) nu mai are si alte solutii.
Tot din (2), (3) si (4) se observa ca (x^2+2)(y^2+3)(z^2+4)>60xyz, daca y>3,\ z>4,\ \forall x\in\mathbb{N}^{\ast}, deci urmeaza ca ecuatia (1) nu are solutii in \mathbb{N}.

Din (1) si (2) \Longrightarrow (y^2+3)(z^2+4)\le 20yz\ (5)

Daca y=1, din (5)\ \Longrightarrow z^2+4\le 5z\ \Longleftrightarrow (z-1)(z-4)\le 0\ \Longleftrightarrow z\in\{1,\ 2,\ 3,\ 4\}. Se arata ca z\in\{2,\ 3\} si y=1 nu conduc la solutii.
Daca y=2, din (5)\ \Longrightarrow 7z^2-40z+28\le 0\ \Longleftrightarrow z\le 5. Verificam in (1) ca z\le 5 si y=2 (nu conduc la solutii).
Daca y=3, din (5)\ \Longrightarrow 12(z^2+4)\le 60z\ \Longleftrightarrow 3z^2-15z+1\le 0\ \Longleftrightarrow 3z(z-5)+1\le 0\ \Longleftrightarrow z\le 4. Aratam ca y=3 si z\in\{2,\ 3\} nu conduc la solutii.
Daca z=1, din (5)\ \Longrightarrow y^2-4y+3\le 0 de unde 1\le y\le 3, caz analizat.
Daca z=2 vom avea y\in\{2,\ 4,\ 5\} care nu dau solutii.
Analog se analizeaza z\in\{3,\ 4\}.
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